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2024 DMC Mock 10 Problems/Problem 20

Problem

If \[a+8b+27c+64d=2\] \[8a+27b+64c+125d=20\] \[27a+64b+125c+216d=202\] \[64a+125b+216c+343d=2024\] then find the remainder when\[125a+216b+343c+512d\]is divided by $1000$.

$\textbf{(A)}\ 222\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 476\qquad\textbf{(D)}\ 938\qquad\textbf{(E)}\ 962$

Solution

Subtract each adjacent pair of equations. This results in: \[7a+19b+37c+61d=18\] \[19a+37b+61c+91d=182\] \[37a+61b+91c+127d=1822\] Repeat this process: \[12a+18b+24c+30d=164\] \[18a+24b+30c+36d=1640\] And again: \[6a+6b+6c+6d=1476\] Next, add this to $18a+24b+30c+36d=1640$, yielding \[24a+30b+36c+42d=3116\] Add the result to $37a+61b+91c+127d=1822$, resulting in \[61a+91b+127c+169d=4938\] Finally, add this equation to $64a+125b+216c+343d=2024$: \[125a+216b+343c+512d=6962\] Thus the answer is $\boxed{\textbf{(E)}\ 962}$.

~ eevee9406

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